[Python-ideas] Include partitioning in itertools

[David Mertz]

I'm surprised to have just noticed that `itertools` is oddly broken this
way (under Python 3.4.3):

This is perfectly happy:

>>> def fibs():

    a, b = 1, 1
    while True:
        yield b
        b, a = a+b, b
...
>>> f = ((a,b) for a in fibs() for b in fibs())
>>> next(f)
Out[3]: (1, 1)
>>> next(f)
Out[4]: (1, 2)
>>> next(f)
Out[5]: (1, 3)

And yet this freezes be greedily trying to go through the infinite iterator:

>>> from itertools import product
>>> product(fibs(), fibs())


On Sun, Nov 1, 2015 at 6:37 PM, Brendan Barnwell <brenbarn at brenbarn.net>
wrote:

> On 2015-11-01 17:58, David Mertz wrote:
>
>> Oh, yeah... I see what you are asking for isn't quite the same thing as
>> power set.  But still, it's not something you can do in finite time with
>> infinite iterators (nor in tractable time with *large* iterators).  So
>> `itertools` isn't where it belongs.
>>
>
>         I don't think that reasoning is sound.  Itertools already includes
> several combinatoric generators like permutations and combinations, which
> aren't computable for infinite iterators and which take intractable time
> for large finite iterators.
>
> --
> Brendan Barnwell
> "Do not follow where the path may lead.  Go, instead, where there is no
> path, and leave a trail."
>    --author unknown
>
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